# Is a continuous and locally Lipschitz function Lipschitz?

Let $f:\Omega \to \mathbb R$ be locally Lipschitz and continuous on $\Omega$, where $\Omega$ is open, bounded subset of $\mathbb R^n.$ From these two conditions can we say that $f$ is Lipschitz?

Edit: It is not true while we are taking $\Omega$ as open , bounded ,but will it be true when $\Omega$ is compact?

• I think $\Omega$ needs to be compact for the result to be true – AmorFati Nov 19 '17 at 11:14
• "Locally Lipschitz and continuous" - doesn't locally Lipschitz already mean continuous? – Mathemagical Nov 19 '17 at 11:51
• Yeah. Now I think that it should have been piecewise Lipschitz and continuous. – nurun nesha Nov 19 '17 at 11:58

No. Take $f\colon(0,1)\longrightarrow\mathbb R$ defined by $f(x)=\frac1x$.

About the new question that you added (you shouldn't; asking a new question is the right way of acting), I could provide an answer, but someone has already done that.

• If we take $\Omega$ to be compact then is this true? – nurun nesha Nov 19 '17 at 11:16
• @mathiu_lady Yes, but then $\Omega$ will not be open. – José Carlos Santos Nov 19 '17 at 11:17
• Yeah of course. Can you give some hints. to prove that – nurun nesha Nov 19 '17 at 11:20
• @mathiu_lady Post it as a question. – José Carlos Santos Nov 19 '17 at 11:21
• Okay. I'm editing my question. Please see it. – nurun nesha Nov 19 '17 at 11:23

Try this: $f:\mathbb{R}_+ \to \mathbb{R}$, defined by $f(x) =\frac{1}{x} .$

• I'd advise not using $\mathbb{R}_+$ as there is no concensus on if the notation includes 0 or not. – JamalS Nov 19 '17 at 11:57
• @JamalS I have never seen it used to include 0 personally. – Wojowu Nov 19 '17 at 16:16